Palladium is an element with symbol Pd  and atomic number 46 (  ie. ₄₆Pd¹⁰⁶ )  . Palladium  , platinum , osmium , iridium , ruthenium , rhodium form a group of metals usually termed as platinum group metals. These have similar chemical  properties ,but palladium has the lowest melting point and is least dense out of them.It has an exceptional electronic configuration  as ,

Electronic configuration of  Pd :  [ Kr ]  4d¹⁰  

Haloarenes  (  also termed as aryl halides )  are class of  aromatic compounds in which one or more number of H atoms directly bonded to an sp² hybridised carbon atom in  an aromatic ring  are replaced by a halide.

Since , the hydrocarbon on complete combustion gives out  given weights CO2  &   H₂O  

Let the weight of hydrocarbon which on combustion gives out  CO₂ and H₂ O  be =  x g

then ,

Step 1 .  /

calculate percentage of H   &   C   present in the hydrocarbon ,  separately

Step 2 /  calculate the number of atoms  of  C  and H  from the % as =   % of element /  atomic mass of element

Step 3 /     Calculate the whole number ratio of C : H in the hydrcarbon

Step 4 /   Infer the empirical formula of the hydrocarbon 

Calculations for Step 1 to 4  are tabulated as below :  

Since empirical formula represents  the simplest whole number ratio of various  present in a compound ,

 {tex}\therefore{/tex}The empirical formula of the hydrocarbon is  C₇ H₈

Following steps / reactions would convert ethanoic acid to  methanoic acid .

1.Reaction  of ethanoic acid with  NH₃  to form  ethanamide ,   C H3 CO  NH₂

CH₃ COOH ---------------NH₃ +  H₂ O----------> CH₃ COONH₄ ---- {tex}\Delta{/tex}---------->  CH₃ CO NH₂   

or, CH₃ COOH    +    NH₃ ----------Al₂O₃ /{tex}\Delta{/tex}-----> CH₃ CO NH₂ 

2.  Treat  ethanamide  with ( alcoholic KOH  +   Br₂ ) to get methanamine CH₃ NH₂  .....(  ie.Hoffmann's Bromamide reaction )

CH₃ CO NH₂  ------  KOH _{(alcoholic)  +}  Br₂   ------------->   CH₃ NH₂     +  K₂ CO₃   +   H₂ O 

3.  Reduction   of  CH₃ NH₂   with  HNO₂  to get methanol  .  

CH₃ NH₂ -------------[H] /  HNO_{2----------->}   CH₃OH   +  N₂    +    H₂ O

4. Oxidation of methanol with ( K₂ Cr₂ O₇ +  dil.  H₂ S O₄) yields  methanoic acid.

CH₃ OH  -------[2 O ] /  ( K ₂ Cr ₂ O₇  +  dil. H ₂SO₄ ) -------- HCOOH     

Calculations :   

Moles of glucose  ,  C₆ H₁₂ O₆ 

=   Mass of glucose  /   malar mass of glucose

=    {tex}\frac{18}{180}{/tex} mol

=  0.1 mol

Number of Kg of solvent  

=  1 kg ........(given)

{tex}\therefore{/tex}molality (m)  of glucose solution 

=   0.1 mol kg^-1 

For water , change in boiling point is given by the expression ,

{tex}\Delta T{/tex}_b    =    K_b x m

where  ,   K_b is Molal boiling point elevation constant is known as  

=  0.52  K  kg mol^-1   

and  m  represents molality

=  0.1 mol kg^-1   ......calculated as above 

So  , substituting the values in above expression we get ,

{tex}\Delta Tb = 0.52kg mol^-1 X 0.1 mol kg^-1{/tex}

=  0.052 K

Since water boils at 373.15 K at 1.013 bar pressure , therefore the boiling point of solution will be 

=  ( 373.15  +   0.052 ) K

=373.202 K